Draw the two diameters AC and BD (in red) 



OA is a radius so OA = 2. And by the marks along the horizontal diameter,
OE is half a radius or half of 2 which is 1.  So OE = 1

So triangle AOE is a 30-60-90 right triangle, so AE=

Triangle AOE has area = 

The four triangles AOE, BOE, COE, DOE are congruent. 
So the four of them have area  

Next we find the area of the sector AOD.
   
Angle AOD is 60o because angles AOE and DOE are both 60o,
so angle AOD = 180o-60o-60o = 60o.

The area of a sector is  
and substituting the values, the area of sector AOD is



The sector BOC is congruent to the sector AOD

Adding the 4 triangles and the two sectors,

Area of ABCD = 

Edwin

I will borrow the plot from the post by Edwin.


    



The area of the figure  ABCD  is the sum of 


      the area of triangle AOB   + the area of triangle DOC + 
    + the area of the sector AOD + the area of the sector BOC.


The area of triangle AOB is the same as the area of equilateral triangle with the side of 2 units,
so it is  =  square units.


The area of triangle AOB + the area of triangle DOC =  square units.


Next, the area of the sector AOD is    of the area of the circle with the radius of 2;
so, the area of the sector AOD is   =  =  square units.


The area of the sector AOD + the area of the sector BOC is twice that value, i.e.    square units.


So, the    is    square units = 3.4641 + 4.1867 = 7.651 square units approximately.